Lecture 20Lecture 20Fluid dynamics.

Bernoulli’s equation.

Rear wingAirplane wing

Rain barrel Tornado damage

Fluid flow

Laminar flow: no mixing between layers

Turbulent flow: a mess…

Dry water, wet water

Real (wet) fluid: friction with walls and between layers (viscosity)

Slower near the walls

Faster in the center

Ideal (dry) fluid: no friction (no viscosity)

Same speed everywhere

Within the case of laminar flow:

Flow rate

Consider a laminar, steady flow of an ideal, incompressible fluid at speed v through a tube of cross-sectional area A

Volume flow rateΔ V

Δ t= A v

A

Δx = v Δt

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Mass flow rateΔ m

Δ t= ρA v

Continuity equation

A1

A2

v1

v2

r1A

1v

1= r

2A

2v

2

The mass flow rate must be the same at any point along the tube (otherwise, fluid would be accumulating or disappearing somewhere)

A1v

1= A

2v

2

If fluid is incompressible (constant density):

ρ1 ρ2

Example: Garden hose

When you use your garden faucet to fill your 3 gallon watering can, it takes 15 seconds. You attach this faucet to your garden hose (radius = 1.5 cm) fitted with a nozzle with 10 holes at the end. Each of the holes is 0.48mm in diameter.

a) What is the speed of water in the hose?

b) What is the speed of water as it spurts through the nozzle?

Example: Garden hose

When you use your garden faucet to fill your 3 gallon watering can, it takes 15 seconds. You attach this faucet to your garden hose (radius = 1.5 cm) fitted with a nozzle with 10 holes at the end. Each of the holes is 0.48mm in diameter.

a) What is the speed of water in the hose?

b) What is the speed of water as it spurts through the nozzle?

a) Volume flow rate: Δ VΔ t

=3gallon

15 s3.785 liter

1 gallon1 m3

1000 liter=7.6×10−4m3/s

vhose =(Δ VΔ t )

Ahose

=7.6×10−4 m3/s

π (1.5×10−2m )2

= 1.1 m/s

b) Volume flow rate is constant (Δ V

Δ t )hose

= (Δ VΔ t )

nozzle

Anozzlevnozzle = 7.6×10−4m3/s

vnozzle =7.6×10−4 m3/s

10π (0.24×10−3m)2= 420 m/s

Δ VΔ t

= Ahosevhose

Work by pressure

A1

A2

v1

v2

As an element of fluid moves during a short interval dt, the ends move distances ds1 and ds2.

ρ1

ρ2

Work by pressure

A1

A2

v1

v2

As an element of fluid moves during a short interval dt, the ends move distances ds1 and ds2.

Work by pressure during its motion: Δ W = p1 A1 Δ s1 − p2 A2 Δ s2 = (p1 − p2) Δ V

ρ1

ρ2

Δs2

Δs1

ΔV

ΔVΔ V = A1 Δ s1 = A2 Δ s2

If the fluid is incompressible, the volume should remain constant:

Kinetic and gravitational potential energy

Change in kinetic energy: Δ KE =1

2(ρ Δ V ) v2

2−

1

2(ρ Δ V ) v1

2

A1

A2

v1

v2

ρ1

ρ2

ds2

ds1

Change in potential energy:

y: height of each element relative to some initial level (eg: floor)

Δ PE = (ρ Δ V) g (y2− y1)

Bernoulli’s equation

Putting everything together: Δ KE + Δ PE = Δ Wnc1

2(ρ Δ V ) v2

2−

1

2(ρ Δ V) v1

2+ (ρ Δ V) (y2−y1) = (p1 − p2) Δ V

2 21 1 1 2 2 2

1 12 2

p v g y p v g yr r r r = 21 constant2

p v g yr r =

For any two points in a continuous flowing incompressible fluid the following equation holds

Static vs flowing fluid

Cylindrical container full of water.

Pressure at point A (hA below surface):

A atm Ap p ghr=

Or gauge pressure: pAgauge = p

A- p

atm= rgh

A

hA

x A

hA

A x

Now we drill a small hole at depth hA.

Point A is now open to the atmosphere! A atmp p=

Container with hole

Assume the radius of the container is R = 15 cm, the radius of the hole is r = 1 cm and hA = 10 cm. How fast does water come out of the hole?

R = 15 cm

hA = 10 cm

yA

yBA x

B x

Bernoulli at points A and B (on the surface):2 2

A A A B B B1 12 2

p v g y p v g yr r r r =

A B atm B A Awhere and p p p y y h= = - =

Continuity at points A and B:

A A B BAv Av=

2 2A B 2 Av v gh- = (Eqn 1)

2 2A Br v R v= (Eqn 2)

2 2A Br v R v=

vB ~ 0 (the container surface moves very slowly because the hole is small ―compared to the container’s base)

vB = ( rR )

2

vA∼0 becauser ≪ R

⇒ vA2∼ 2ghA

R = 15 cm

hA = 10 cm

yA

Speed at the hole, is same as speed of free fall from same height!!!“Torricelli's Theorem” ~100 year before Bernoulli's equation

v = √2ghA

2A 2 2 9.8 m/s 0.10 m 1.4 m/sAv gh= = =

v = √2ghA

h

●A ●B

flow

Measuring fluid speed: the Venturi meter

A horizontal pipe of radius RA carrying water has narrow throat of radius RB. Two small vertical pipes at points A and B show a difference in level of h. Measuring the pressure at points A and B, gives the speeds At these points.

2 2A A B B A B

Bernouilli:1 1 2 2

p v p v y yr r = =

A B

Statics:p p ghr- =

Venturi effect:High speed, low pressureLow speed, high pressure

2 equations for vA, vB

Continuity:AA vA=ABvB ⇒ RA

2 vA=RB2 vB

2 2B A A B

1 2

v v p pr - = -

2 2A A B B

2 2A A B B

1 12 2

p v p v

R v R v

r r =

=

A Band p p ghr- =

vB = (RA

RB)2

vA

DEMO:

Tube with changing diameter

12

ρ[(RA

RB)2

−1]vA = pA−pB

vA =

√2gh

(RA

RB)2

−1

Partially illegal Bernoulli

Gases are NOT incompressible

Bernoulli’s equation cannot be used

It can be used if the speed of the gas is not too large (compared to the speed of sound in that gas).

But…

i.e., if the changes in density are small along the streamline

Example: Why do planes fly?

High speed, low pressure

Low speed, high pressure

Net force up (“Lift”)

Bernoulli: p

top

12rv

top2 = p

bottom

12rv

bottom

2 rgDh is negligible

bottom

2 2top topbottomLift area of wing area of wing

2p p v vr

= - = -

DEMO:

Paper sucked by blower.

DEMO:

Beach ball

trapped in air.

Aerodynamic grip

Tight space under the car fast moving air low pressure➝ ➝

Race cars use the same effect in opposite direction to increase their grip to the road (important to increase maximum static friction to be able to take curves fast)

Lower pressure

Higher pressure

Net force down

ACT: Blowing across a U-tube

A U-tube is partially filled with water. A person blows across the top of one arm. The water in that arm:

A. Rises slightly

B. Drops slightly

C. It depends on how hard is the blowing.

Tornadoes and hurricanes

Strong winds Low pressures➝

vin = 0vout = 250 mph (112 m/s)

Upward force on a 10 m x 10 m roof: F = (7500 Pa) (10 m)2

= 7.5×105 N

Weight of a 10 m x 10 m roof (0.1 m thick and using density of water –wood is lighter than water but all metal parts are denser):

4 2 510 kg 10 m/s 10 Nmg = =

The roof is pushed off by the air inside !

pin−pout =12

ρvout2

=12

(1.2kg/m3) (112m/s)2= 7500Pa

The suicide door

The high speed wind will also push objects when the wind hits a surface perpendicularly!

Air pressure decreases due to air moving along a surface.

Modern car doors are never hinged on the rear side anymore.

If you open this door while the car is moving fast, the pressure difference between the inside and the outside will push the door wide open in a violent movement.

In modern cars, the air hits the open door and closes it again.

Curveballs

Speed of air layer close to ball is reduced (relative to ball)

Boomerangs are based on the same principle (Magnus effect)

Speed of air layer close to ball is increased (relative to ball)

Beyond Bernoulli

In the presence of viscosity, pressure may decrease without an increase in speed.

Example: Punctured hose (with steady flow).

Speed must remain constant along hose due to continuity equation.

Ideal fluid (no viscosity) Real fluid (with viscosity)Friction accounts for the decrease in pressure.

Lower jet.

The syphon

The trick to empty a clogged sink:

A x

x B

h

Bernoulli:

pA

12rv

A2 rgh = p

B

12rv

B2

Thin hose → vA ~ 0

=B 2v gh

PA = PB = Patm